How to calculate tension in physics. Weightless block thread tension friction acceleration mass load find force

In physics, tension is the force acting on a rope, cord, cable or similar object or group of objects. Anything that is pulled, suspended, supported, or swings by a rope, cord, cable, etc., is the object of a tension force. Like all forces, tension can accelerate objects or cause them to deform. The ability to calculate tensile force is an important skill not only for students of the Faculty of Physics, but also for engineers and architects; those who build stable homes need to know whether a particular rope or cable will withstand the tension force of the object's weight without sagging or collapsing. Start reading this article to learn how to calculate the tension force in some physical systems.

Steps

Determination of tension on one thread

1. Determine the forces at each end of the thread. The tension in a given thread or rope is the result of forces pulling on the rope at each end. We remind you that force = mass × acceleration. Assuming the rope is taut, any change in the acceleration or mass of an object suspended from the rope will result in a change in the tension force in the rope itself. Don't forget about the constant acceleration of gravity - even if the system is at rest, its components are subject to gravity. We can assume that the tension force of a given rope is T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any of the objects supported by the rope, and "a" is any other acceleration, acting on objects.

   • To solve many physical problems, we assume perfect rope- in other words, our rope is thin, has no mass and cannot stretch or break.
   • As an example, let's consider a system in which a load is suspended from a wooden beam using a single rope (see image). Neither the load itself nor the rope moves - the system is at rest. As a result, we know that in order for the load to be in equilibrium, the tension force must be equal to the force of gravity. In other words, Tension (F t) = Gravity (F g) = m × g.
     • Let's assume that the load has a mass of 10 kg, therefore the tension force is 10 kg × 9.8 m/s 2 = 98 Newtons.

2. Consider acceleration. Gravity is not the only force that can affect the tension of a rope - the same effect is produced by any force applied to an object on a rope with acceleration. If, for example, an object suspended from a rope or cable is accelerated by a force, then the acceleration force (mass × acceleration) is added to the tension force generated by the weight of the object.

   • In our example, suppose that a 10 kg load is suspended from a rope and, instead of being attached to a wooden beam, it is pulled upward with an acceleration of 1 m/s 2 . In this case, we need to take into account the acceleration of the load as well as the acceleration of gravity, as follows:
     • F t = F g + m × a
     • F t = 98 + 10 kg × 1 m/s 2
     • F t = 108 Newtons.

3. Consider angular acceleration. An object on a rope rotating about a point considered the center (like a pendulum) exerts tension on the rope through centrifugal force. Centrifugal force is the additional tension force caused by the rope, "pushing" it inward so that the load continues to move in an arc rather than in a straight line. The faster an object moves, the greater the centrifugal force. Centrifugal force (F c) is equal to m × v 2 /r where “m” is the mass, “v” is the speed, and “r” is the radius of the circle along which the load is moving.

   • Since the direction and magnitude of centrifugal force changes depending on how the object moves and changes its speed, the total tension in the rope is always parallel to the rope at the center point. Remember that the force of gravity is constantly acting on an object and pulling it down. So if the object is swinging vertically, the full tension strongest at the bottom of the arc (for a pendulum this is called the equilibrium point) when the object reaches its maximum speed, and weakest at the top of the arc as the object slows down.
   • Let's assume that in our example the object is no longer accelerating upward, but is swinging like a pendulum. Let our rope be 1.5 m long, and our load move at a speed of 2 m/s when passing through the lower point of the swing. If we need to calculate the tension force at the bottom point of the arc, when it is greatest, then we first need to find out whether the pressure of gravity is experienced by the load at this point, as at rest - 98 Newtons. To find the additional centrifugal force, we need to solve the following:
     • F c = m × v 2 /r
     • F c = 10 × 2 2 /1.5
     • F c =10 × 2.67 = 26.7 Newtons.
     • So the total tension will be 98 + 26.7 = 124.7 Newton.

4. Please note that the tension force due to gravity changes as the load passes through the arc. As noted above, the direction and magnitude of centrifugal force changes as the object swings. In any case, although gravity remains constant, net tension force due to gravity is also changing. When the swinging object is Not at the bottom of the arc (equilibrium point), gravity pulls it down, but tension pulls it up at an angle. For this reason, the tension force must counteract part of the force of gravity, not all of it.

   • Dividing the force of gravity into two vectors can help you visualize this state. At any point in the arc of a vertically swinging object, the rope makes an angle "θ" with a line passing through the equilibrium point and the center of rotation. As soon as the pendulum begins to swing, the gravitational force (m × g) is divided into 2 vectors - mgsin(θ), acting tangentially to the arc in the direction of the equilibrium point and mgcos(θ), acting parallel to the tension force, but in the opposite direction. Tension can only resist mgcos(θ) - the force directed against it - not the entire force of gravity (except at the equilibrium point, where all forces are equal).
   • Let's assume that when the pendulum is tilted at an angle of 15 degrees from the vertical, it moves at a speed of 1.5 m/s. We will find the tension force by the following steps:
     • Ratio of tension force to gravitational force (T g) = 98cos(15) = 98(0.96) = 94.08 Newton
     • Centrifugal force (F c) = 10 × 1.5 2 /1.5 = 10 × 1.5 = 15 Newtons
     • Total tension = T g + F c = 94.08 + 15 = 109.08 Newtons.

5. Calculate the friction. Any object that is pulled by a rope and experiences a "braking" force from the friction of another object (or fluid) transfers this force to the tension in the rope. The friction force between two objects is calculated in the same way as in any other situation - using the following equation: Friction force (usually written as F r) = (mu)N, where mu is the coefficient of friction force between objects and N is the usual force of interaction between objects, or the force with which they press on each other. Note that static friction, which is the friction that results from trying to force an object at rest into motion, is different from motion friction, which is the friction that results from trying to force a moving object to continue moving.

   • Let's assume that our 10 kg load is no longer swinging, but is now being towed along a horizontal plane using a rope. Let's assume that the coefficient of friction of the earth's motion is 0.5 and our load is moving at a constant speed, but we need to give it an acceleration of 1 m/s 2 . This problem introduces two important changes - first, we no longer need to calculate the tension force in relation to gravity, since our rope is not holding a weight suspended. Second, we will have to calculate the tension due to friction as well as that due to the acceleration of the mass of the load. We need to decide the following:
     • Normal force (N) = 10 kg & × 9.8 (gravity acceleration) = 98 N
     • Motion friction force (F r) = 0.5 × 98 N = 49 Newtons
     • Acceleration force (F a) = 10 kg × 1 m/s 2 = 10 Newton
     • Total tension = F r + F a = 49 + 10 = 59 Newtons.

   Calculation of tension force on several threads

   1. Lift vertical parallel weights using a block. Pulleys are simple mechanisms consisting of a suspended disk that allows you to change the direction of the tension force on the rope. In a simple pulley configuration, a rope or cable runs from a suspended weight up to a pulley, then down to another weight, thereby creating two sections of rope or cable. In any case, the tension in each of the sections will be the same, even if both ends are tensioned by forces of different magnitudes. For a system of two masses suspended vertically in a block, the tension force is equal to 2g(m 1)(m 2)/(m 2 +m 1), where “g” is the acceleration of gravity, “m 1” is the mass of the first object, “ m 2 ” – mass of the second object.

      • Note the following: physical problems assume that the blocks are perfect- have no mass, no friction, they do not break, are not deformed and do not separate from the rope that supports them.
      • Let's assume that we have two weights suspended vertically at parallel ends of a rope. One weight has a mass of 10 kg, and the second has a mass of 5 kg. In this case, we need to calculate the following:
        • T = 2g(m 1)(m 2)/(m 2 +m 1)
        • T = 2(9.8)(10)(5)/(5 + 10)
        • T = 19.6(50)/(15)
        • T = 980/15
        • T= 65.33 Newtons.
      • Note that since one weight is heavier, all other elements are equal, this system will begin to accelerate, hence the 10 kg weight will move down, causing the second weight to go up.

   2. Hang weights using pulleys with non-parallel vertical strings. Blocks are often used to direct the tension force in a direction other than down or up. If, for example, a load is suspended vertically from one end of a rope, and the other end holds the load in a diagonal plane, then the non-parallel system of pulleys takes the shape of a triangle with corners at the points of the first load, the second and the pulley itself. In this case, the tension in the rope depends both on gravity and on the component of the tension force that is parallel to the diagonal part of the rope.

      • Let's assume that we have a system with a 10 kg (m 1) load suspended vertically, connected to a 5 kg (m 2) load placed on a 60 degree inclined plane (this inclination is assumed to be frictionless). To find the tension in a rope, the easiest way is to first set up equations for the forces accelerating the loads. Next we proceed like this:
        • The suspended weight is heavier, there is no friction, so we know it is accelerating downward. The tension in the rope pulls upward, so that it accelerates with respect to the resultant force F = m 1 (g) - T, or 10(9.8) - T = 98 - T.
        • We know that a mass on an inclined plane accelerates upward. Since it has no friction, we know that tension pulls the load up along the plane, and pulls it down only your own weight. The component of the force pulling down the slope is calculated as mgsin(θ), so in our case we can conclude that it is accelerating with respect to the resultant force F = T - m 2 (g)sin(60) = T - 5( 9.8)(0.87) = T - 42.14.
        • If we equate these two equations, we get 98 - T = T - 42.14. We find T and get 2T = 140.14, or T = 70.07 Newtons.

   3. Use multiple strings to hang the object. Finally, let's imagine that the object is suspended from a "Y-shaped" system of ropes - two ropes are fixed to the ceiling and meet at a central point from which a third rope with a weight extends. The tension on the third rope is obvious - simple tension due to gravity or m(g). The tensions on the other two ropes are different and must add up to a force equal to the force of gravity upward in the vertical position and zero in both horizontal directions, assuming the system is at rest. The tension in a rope depends on the mass of the suspended loads and on the angle at which each rope is tilted from the ceiling.

      • Let's assume that in our Y-shaped system the bottom weight has a mass of 10 kg and is suspended on two ropes, one of which makes an angle of 30 degrees with the ceiling, and the second of which makes an angle of 60 degrees. If we need to find the tension in each of the ropes, we will need to calculate the horizontal and vertical components of the tension. To find T 1 (tension in the rope whose inclination is 30 degrees) and T 2 (tension in that rope whose inclination is 60 degrees), you need to solve:
        • According to the laws of trigonometry, the ratio between T = m(g) and T 1 and T 2 is equal to the cosine of the angle between each of the ropes and the ceiling. For T 1, cos(30) = 0.87, as for T 2, cos(60) = 0.5
        • Multiply the tension in the bottom rope (T=mg) by the cosine of each angle to find T 1 and T 2 .
        • T 1 = 0.87 × m(g) = 0.87 × 10(9.8) = 85.26 Newtons.
        • T 2 =0.5 × m(g) = 0.5 × 10(9.8) = 49 Newtons.

A thread is thrown through a weightless block, connecting body 3 with body 2, from which body 1 is suspended. The mass of each body is 2 kg. Find the acceleration of body 1 and the tension force of the thread connecting it to body 2.

problem 12431

In the installation (Fig. 3), the angle α = 50° of the inclined plane with the horizon of body mass m 1 = 0.15 kg and m 2 = 0.5 kg. Assuming the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if a body of mass m2 is lowered.

problem 13039

Two loads ( m 1 = 500 g and m 2 = 700 g) are tied with a weightless thread and lie on a smooth horizontal surface. To cargo m 1 horizontally directed force is applied F= 6 N. Neglecting friction, determine 1) the acceleration of the loads; 2) the tension of the thread.

task 13040

The simplest Atwood machine, used to study the laws of uniformly accelerated motion, consists of two loads with unequal masses m 1 and m 2 (for example m 1 > m 2), which are suspended on a light thread thrown over a stationary block. Assuming the thread and the block to be weightless and neglecting friction in the axis of the block, determine 1) the acceleration of the loads; 2) thread tension T; 3) strength F, acting on the axis of the block.

problem 13041

Loads with masses m 1 = 200 g and m 2 = 500 g are suspended from a system of blocks (see figure). The load m 1 rises, the movable block with m 2 lowers, the blocks and thread are weightless, there are no friction forces. Determine: 1) the tension force of the thread T; 2) acceleration of cargo.

problem 13042

Bodies with masses m 1 = 200 g and m 2 = 150 g are connected by a weightless thread. The angle α between the inclined plane and the horizon is 20°. Neglecting the friction forces and considering the block weightless, determine the acceleration with which the bodies move, assuming that the body m 2 is moving down.

task 13043

On a horizontal table there is a body A with mass M = 2 kg, connected by threads with blocks to bodies B (m 1 = 0.5 kg) and C (m 2 = 0.3 kg). Considering blocks and threads weightless and neglecting friction forces, find: 1) the acceleration with which these bodies move; 2) the difference in thread tension forces.

task 13044

The angles between the inclined planes and the horizon are specified: α=30° and β=45°. A weightless thread connecting bodies with masses m 1 = 0.45 kg and m 2 = 0.5 kg is thrown over a weightless block. Find: 1) acceleration of body motion; 2) thread tension force. Neglect frictional forces.

task 13052

A load lying on the table is connected by a thread thrown over a weightless block on the edge of the table with a hanging load of the same mass (m 1 = m 2 = 0.5 kg). The coefficient of friction of the load m 2 on the table is f = 0.15. Find: 1) acceleration of loads; 2) the tension of the thread. Neglect block friction.

task 13055

The angle α between the plane and the horizon is 30°, the masses of the bodies are identical in m = 1 kg. A body lies on a plane, the friction coefficient between which and the plane is f = 0.1. Neglecting friction in the axis of the block and considering the block and thread to be weightless, determine the force of pressure on the axis.

task 13146

A weightless thread, at the ends of which bodies with masses m 1 = 0.35 kg and m 2 = 0.55 kg are attached, is thrown through a stationary block in the form of a solid homogeneous cylinder with mass m = 0.2 kg. Find: 1) acceleration of loads; 2) the ratio T 2 /T 1 of the tension forces of the threads. Neglect friction in the block axis.

task 13147

Using a block in the form of a thin-walled hollow cylinder, a body of mass m 1 = 0.25 kg is connected by a weightless thread to a body of mass m 2 = 0.2 kg. The first body slides along the surface of a horizontal table with a friction coefficient f equal to 0.2. Block mass m = 0.15 kg. Neglecting friction in the bearings, determine: 1) acceleration a of the bodies; 2) tension forces T 1 and T 2 of the thread on both sides of the block.

task 14495

Two weights with masses m 1 = 2 kg and m 2 = 1 kg are connected by a thread and thrown over a weightless block. Find the acceleration a with which the weights move and the tension force of the thread T. Neglect friction in the block.

task 14497

A weightless block is fixed at the top of an inclined plane making an angle α = 30° with the horizon. Weights 1 and 2 of the same mass m 1 = m 2 = 1 kg are connected by a thread and thrown over a block. Find the acceleration a with which the weights move and the tension force of the thread T. Neglect the friction of the weight on the inclined plane and the friction in the block.

task 14499

A weightless block was strengthened on the top of two inclined planes, which made angles α = 30° and β = 45° with the horizon, respectively. Weights 1 and 2 of the same mass m 1 = m 2 = 1 kg were connected by a thread thrown over a block. Find the acceleration a with which the weights move and the tension force of the thread T. The friction of the weights on inclined planes and the friction in the block can be neglected.

task 15783

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 = 0.5 kg and m 2 = 0.2 kg, which are suspended on a light thread thrown over a stationary block. Considering the thread and the block weightless and neglecting friction in the axis of the block, determine: 1) acceleration of the loads; 2) the tension of the thread.

task 15785

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 = 0.6 kg and m 2 = 0.2 kg, which are suspended on a light thread thrown over a stationary block. Considering the thread and the block weightless and neglecting friction in the axis of the block, determine: 1) acceleration of the loads; 2) the tension of the thread.

task 15787

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 = 0.8 kg and m 2 = 0.15 kg, which are suspended on a light thread thrown over a stationary block. Considering the thread and the block weightless and neglecting friction in the axis of the block, determine: 1) acceleration of the loads; 2) the tension of the thread.

problem 15789

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 = 0.35 kg and m 2 = 0.55 kg, which are suspended on a light thread thrown over a stationary block. Considering the thread and the block weightless and neglecting friction in the axis of the block, determine: 1) acceleration of the loads; 2) the tension of the thread.

task 15791

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 = 0.8 kg and m 2 = 0.2 kg, which are suspended on a light thread thrown over a stationary block. Considering the thread and the block weightless and neglecting friction in the axis of the block, determine: 1) acceleration of the loads; 2) the tension of the thread.

problem 15796

In the installation (Fig. 3), the angle α = 30° of the inclined plane with the horizon of body masses m 1 = 300 g and m 2 = 0.8 kg. Assuming the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if a body of mass m2 is lowered.

task 15798

In the installation (Fig. 3), the angle α = 60° of the inclined plane with the horizon of body mass m 1 = 500 g and m 2 = 0.6 kg. Assuming the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if a body of mass m2 is lowered.

task 15800

In the installation (Fig. 3), the angle α = 20° of the inclined plane with the horizon of body mass m 1 = 350 g and m 2 = 0.2 kg. Assuming the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if a body of mass m2 is lowered.

task 15802

In the installation (Fig. 3), the angle α = 60° of the inclined plane with the horizon of body masses m 1 = 100 g and m 2 = 0.2 kg. Assuming the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if a body of mass m2 is lowered.

problem 17126

In the installation (Fig. 2.13), the angles α and β with the horizon are respectively equal to 45° and 30°, masses of bodies m 1 = 0.5 kg and m 2 = 0.45 kg. Considering the thread and the block weightless and neglecting the forces of friction, determine: 1) the acceleration with which the bodies move; 2) the tension of the thread.

problem 17211

Bodies with masses m 1 = 5 kg and m 2 = 3 kg are connected by a weightless thread thrown through a block of mass m = 2 kg and radius r = 10 cm; they lie on conjugate inclined planes with inclination angles β = 30°. Body m 2 is acted upon by a vertical force F equal to 15

task 40125

Weights of the same mass (m 1 = m 2 = 0.5 kg) are connected by a thread and thrown over a weightless block mounted at the end of the table. The coefficient of friction of the load m 2 on the table is µ = 0.15. Neglecting friction in the block, determine: a) the acceleration with which the loads move; b) the tension of the thread.

task 40126

A weightless thread is thrown through a block in the form of a homogeneous disk with a mass of 80 g, to the ends of which are attached loads with masses m 1 = 100 g and m 2 = 200 g. Find the acceleration with which the loads will move? Ignore friction.

task 40482

Two different weights are attached to the ends of a weightless thread thrown over a block of radius 0.4 m with a moment of inertia of 0.2 kg m 2. The moment of friction forces when the block rotates is 4 Nm. Find the difference in tension in the thread on both sides of the block rotating with a constant angular acceleration of 2.5 rad/s 2 .

task 40499

At the top of two inclined planes forming angles α = 28° and β = 40° with the horizon, a block is fixed. Weights with equal masses are attached to a thread thrown over a block. Assuming the thread and block to be weightless and neglecting friction, determine the acceleration a of the loads.

task 40602

Attached to the ceiling of an elevator descending with acceleration a l is the free end of a thin and weightless thread, which is wound around a thin-walled hollow cylinder of mass m. Find the acceleration of the cylinder relative to the elevator and the tension in the string. Consider the thread vertical.

task 40620

Weights of masses 19 kg and 10 kg were connected by a thread thrown through a weightless block fixed to the ceiling. Neglecting friction in the block, determine the tension of the thread.

task 40623

An inclined plane, on the top of which a weightless block is fixed, forms an angle of 19 degrees with the horizon. Two weights of equal mass 5 kg are attached to the ends of a thread thrown over a pulley. In this case, one of the weights moves along an inclined plane, and the other hangs vertically on a thread, without touching the plane. Find the thread tension. Neglect friction in the block and friction on the plane.

Movement of a system of bodies

Dynamics: motion of a system of connected bodies.

Projection of forces of several objects.

The action of Newton's second law on bodies that are held together by a thread

If you, my friend, have forgotten how to project, I advise you to refresh your little head.

And for those who remember everything, let's go!

Problem 1. On a smooth table lie two bars connected by a weightless and inextensible thread with a mass of 200 g on the left and a mass on the right 300 g. A force of 0.1 N is applied to the first, and a force of 0.6 N is applied to the left in the opposite direction. With what acceleration are they moving? cargo?

Movement occurs only on the X axis.

Because If a large force is applied to the right load, the movement of this system will be directed to the right, so we will direct the axis in the same way. The acceleration of both bars will be directed in one direction - the side of greater force.

Let's add the upper and lower equations. In all problems, unless there are some conditions, the tension force of different bodies is the same T₁ and T₂.

Let's express the acceleration:

Task 2. Two bars connected by an inextensible thread are located on a horizontal plane. Forces F₁ and F₂ are applied to them, making angles α and β with the horizon. Find the acceleration of the system and the tension in the thread. The coefficients of friction between the bars and the plane are the same and equal to μ. The forces F₁ and F₂ are less than the gravity force of the bars. The system moves to the left.

The system moves to the left, but the axis can be directed in any direction (it’s just a matter of signs, you can experiment at your leisure). For a change, let's point to the right, against the movement of the entire system, we love minuses! Let's project forces onto Oh (if there are difficulties with this -).

According to II. Newton, we project the forces of both bodies onto Ox:

Let's add up the equations and express the acceleration:

Let us express the tension of the thread. To do this, we equate the acceleration from both equations of the system:

Task 3. A thread is thrown through a stationary block, from which three identical weights (two on one side and one on the other) with a mass of 5 are suspended.kg each. Find the acceleration of the system. How far will the loads travel in the first 4 seconds of movement?

In this problem, we can imagine that the two left weights are fastened together without a thread, this will save us from projecting mutually equal forces.

Subtract the second from the first equation:

Knowing the acceleration and the fact that the initial speed is zero, we use the path formula for uniformly accelerated motion:

Problem 4. Two masses of masses 4 kg and 6 kg are connected by a light inextensible thread. Friction coefficients between load and tableμ = 0.2. Determine the acceleration with which the loads will move.

Let us write down the motion of the bodies on the axis, and from Oy we find N for the friction force (Ftr = μN):

(If it is difficult to understand what equations will be needed to solve the problem, it is better to write everything down)

Let's add the two lower equations so that T is reduced:

Let's express the acceleration:

Task 5. A block of mass 6 kg lies on an inclined surface with an angle of inclination of 45°. A mass of 4 kg is attached to a block using a thread and thrown over the block. Determine the tension of the thread if the coefficient of friction of the bar on the plane is μ = 0.02. At what values ​​of μ will the system be in equilibrium?

Let's direct the axis arbitrarily and assume that the right load outweighs the left one and lifts it up the inclined plane.

From the equation for the Y axis, we express N for the friction force on the X axis (Ftr = μN):

Let's solve the system by taking the equation for the left body along the X axis and for the right body along the Y axis:

Let's express the acceleration so that there is only one unknown T left, and find it:

The system will be in equilibrium. This means that the sum of all forces acting on each of the bodies will be equal to zero:

If you received a negative coefficient of friction, it means that you chose the movement of the system incorrectly (acceleration, friction force). You can check this by substituting the tension force of the thread T into any equation and finding the acceleration. But it’s okay, the values ​​remain the same in magnitude, but opposite in direction.

This means that the correct direction of the forces should look like this, and the coefficient of friction at which the system will be in equilibrium is equal to 0.06.

Problem 6. On two inclined planes there is a load with masses of 1 kg. The angle between the horizontal and the planes is α= 45° and β = 30°. Friction coefficient for both planes μ= 0.1. Find the acceleration with which the weights move and the tension in the string. What should be the ratio of the masses of the loads so that they are in equilibrium.

This problem will require all the equations on both axes for each body:

Let's find N in both cases, substitute them for the force of friction and write together the equations for the X axis of both bodies:

Let's add up the equations and reduce by mass:

Let's express the acceleration:

Substituting the found acceleration into any equation, we find T:

Now let’s overcome the last point and figure out the mass ratio. The sum of all forces acting on any of the bodies is equal to zero in order for the system to be in equilibrium:

Let's add up the equations

Let’s move everything that has the same mass into one part, and everything else into the other part of the equation:

We found that the mass ratio should be as follows:

However, if we assume that the system can move in a different direction, that is, the right load will outweigh the left one, the direction of acceleration and friction force will change. The equations will remain the same, but the signs will be different, and then the mass ratio will be like this:

Then, with a mass ratio from 1.08 to 1.88, the system will be at rest.

Many may have the impression that the mass ratio should be some specific value, and not a gap. This is true if there is no frictional force. To balance the forces of gravity at different angles, there is only one option when the system is at rest.

In this case, the friction force gives a range in which, until the friction force is overcome, movement will not begin.